I'm not sure why the fermi surface crosses the Brillouin zone boundary at right angles. I understand that this is normally the case, but not necessarily always.

I'm aware that the fermi surface is a constant energy surface up to the filling point. The Brillouin zone is in reciprocal space.


This answer nothing to the OP's question, please don't vote up anymore; and anyone who know the answer to this question please share.

This is mainly due to the time reversal symmetry.

Consider the Bloch equation:


Recall that $\psi_{nk}=e^{ik\cdot r}u_{nk}$, then we have:


Now we want to prove $\epsilon_{nk}=\epsilon_{n-k}$, take the complex conjugate of the above equation and change $k\to-k$:


We can see that $\epsilon_{n-k}$ is the same set of eigenvalues as $\epsilon_{nk}$ of the same Hamiltonian $H_k$. Thus they must be equal.

Now let's answer your question:

Consider one particular band $n_0$, its zone boundary are $-K/2$ and $K/2$.

$\epsilon_{n_0,K/2+\Delta k}=\epsilon_{n_0,-K/2+\Delta > k}=\epsilon_{n_0,K/2-\Delta k}$

Let's $\Delta k$ tends to infinity small, the above equation just means that the first derivative of energy band near the zone boundary is zero.

So that when the filling of electrons doesn't modify the band structure, you will always see the fermi surface perpendicular to the zone boundary if time reversal symmetry is respected.

  • $\begingroup$ This merely proves that time reversal symmetry renders the BZ boundaries as isoenergetic surfaces, which doesn't answer the question. $\endgroup$ – xiaohuamao Sep 30 '14 at 14:23

I believe it has to do with translation, inversion, and mirror symmetry?

In real space, the potential energy has to share the same periodicity as the lattice. Similarly in k-space, the Fermi energy surface has to share the same periodicity as the reciprocal lattice. Brillouin zone boundaries are like the markers in k-space that specify when one period ends and the next begins. Periodicity alone doesn't guarantee that the Fermi surface cuts the zone boundaries at 90 degrees. But with inversion and mirror symmetries it does! Not all crystal structures have inversion and mirror symmetries. But the common ones like BCC and FCC do.


I would like to argue that the result follows if one demands that energy is differentiable in k space. To be precise, one would need $E = E\,(k_x, k_y)$ such that $\nabla_{\mathbf{k}}E$ always exists (I have reduced the dimensionality for ease of visualisation).

This is perhaps essential on physical grounds, since the group velocity of electron wavepackets is $\frac{1}{\hbar}\nabla_{\mathbf{k}}E$. Imagine forcing an electron along in k space using some external force. Suppose that $\nabla_{\mathbf{k}}E$ took different values on two directions of approach to the same point. As the electron's path crosses this point, its group velocity would have experienced a discontinuous jump, which is arguably unphysical since only an infinitesimal impulse was applied in getting it from one side of the discontinuity to the other. (This is only a plausability argument that will hopefully be convincing enough...)

Getting back to the main argument, the Fermi surface in two dimensions is a curve of constant energy (extending it in the third dimension makes a sheet). This implies that any vector $\mathbf{\delta k}$ that is tangent to the curve must satisfy $\mathbf{\delta k}\cdot\nabla_{\mathbf{k}}E = 0$. In other words, $\nabla_{\mathbf{k}}E$ is normal to the curve at any point. If the Fermi curve were to intersect a Brillouin zone boundary at anything other than a right angle, reflecting the curve across the boundary (by lattice periodicity) would produce a kink at that boundary. One would then find that the Fermi curve has two different normal vectors when approaching the boundary from different zones. Hence $\nabla_{\mathbf{k}}E$ does not exist at the zone boundary, which violates our initial requirement.

There is one important exception when $\nabla_{\mathbf{k}}E$ is exactly zero at the point where the Fermi curve intersects the zone boundary. Then $\mathbf{\delta k}\cdot\nabla_{\mathbf{k}}E = 0$ is trivially satisfied for arbitrary $\mathbf{\delta k}$, and the Fermi curve is allowed to approach this point from any angle. This of course corresponds to the case where there is a stationary point at the zone boundary.


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